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3 years ago

When y⁷ is multiplied by y⁹ the answer is?

Physics

1 answer:

Allushta [10]3 years ago

5 0

Answer:

y^16

Explanation:

who need to add the exponents only

7 + 9 = 16

therefore, the answer is y^16

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A 0.900 kg block is attached to a spring with spring constant 14.5 N/m . While the block is sitting at rest, a student hits it w

olga55 [171]

Answer:

1. $A=0.0847\ m$

2. $v=0.050\ m.s^{-1}$

Explanation:

Given:

mass of block, $m=0.9\ kg$

spring constant, $k=14.5\ N.m^{-1}$

maximum velocity of block, $v_m=34\ cm.s^{-1}$

<u>We know from the energy of oscillating spring:</u>

$E=\frac{1}{2}.k.A^2= \frac{1}{2}.m.v_m^2$

$\frac{1}{2}\times 14.5\times A^2=\frac{1}{2}\times 0.9\times 0.34^2$

$A=0.0847\ m$

Now, given that:

instantaneous position, $x=0.15\times A=0.0127\ m$

we find angular speed,

$\omega=\sqrt{\frac{k}{m} }$

$\omega=\sqrt{\frac{14.5}{0.9} }$

$\omega=4.013\ rad.s^{-1}$

So we have

$v=x.\omega$

$v=0.0127\times 4.013$

$v=0.051\ m.s^{-1}$

7 0

4 years ago

What is the gpe of a 200 kg hot air ballon 21,000 m above the ground?

stiv31 [10]

gravitational Potential energy is given by

$GPE = mgh$

here we have

m = 200 kg

h = 21000 m

now we will have

$GPE = 200(9.8)(21000)$

$GPE = 4.116 \times 10^7 J$

so GPE of balloon will be 41160000 J above the given height from ground

8 0

3 years ago

What is the difference between mass and weight?

xxTIMURxx [149]

Answer:

C. Mass is independent of the force acting on the mass.

Weight depends on the gravitational force acting on the mass.

Example: A mass of 1 kg on the earth still has a mass of 1 kg on the moon;

whereas the weight of that mass on earth is 9.8 N and only 1/6 of that amount on the moon.

4 0

4 years ago

During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 6.82 s, how high does it rise? The accel

scoray [572]

The answer is 56.98 m

The dislocation (d) is:
d = v1 * t + 1/2 * a * t²
v1 - initial velocity
t - time
a - <span>acceleration of gravity

We know:
d = ?
v1 = ?
t = 6.82 s / 2 = 3.41 (it reaches the peak at half time)
a = - 9.8 m/s</span>²

Let's first calculate v1:
v2 = v1 + at
v2 - final velocity (v2 = 0 when it reaches peak)

0 = v1 + -9.8 * 3.41
0 = v1 - 33.418

v1 = 33.418 m/s

d = v1 * t + 1/2 * a * t²
d = 33.418 * 3.41 + 1/2 * -9.8 * 3.41²
d = 113.96 - 56.98
d = 56.98 m

8 0

4 years ago

Read each scenario and then answer the questions. Scenario A: A shopping cart is pushed a distance of 3 m by a force of 15 N. Sc

ValentinkaMS [17]

To solve this we are going to use the formula for work: $W=Fd$
where
$W$ is the work done
$F$ is the force
$d$ is the distance

Scenario A. We know for our problem that $F=15N$ and $d=3m$ . Lets replace those values in our formula to find $W$ :
$W=15N*3m$
$W=45J$

Scenario B. We know for our problem that $F=12N$ and $d=4$ . Lest replace those value in our formula:
$W=12N*4m$
$W=48J$

Scenario C. We know for our problem that $F=10N$ and $d=6m$ . Lets replace the values in our formula:
$F=10N*6m$
$F=60J$

We can conclude that:
- The scenario that requires the most work is C.
- The scenario that requires the least work is A.

7 0

3 years ago

Read 2 more answers